Thread: Inflection Point Fun

1. Inflection Point Fun

More inflection pt fun!! No idea where to start on this one. Any help is greatly appreciated:

g(x) = e^[(-(x-a)^2)/(2b^2)] / sqrt[2*pi*b^2]

2. $g(x) = \frac{e^{\frac{-(x-a)^2}{2b^2}}}{\sqrt{2 \pi b^2}}$

Differentiate twice, i.e find $g''(x)$ and solve $g''(x)=0$ to find any inflection points.

3. Can anyone find the 1st and 2nd derivatives of this problem and solve for 0 to find inflection pts?

4. Originally Posted by mat220-09
Can anyone find the 1st and 2nd derivatives of this problem and solve for 0 to find inflection pts?

You mean "Can anyone do this entire question for me?".

That's not how it works.

To get the first derivative you have to use the chain rule. Where are you stuck in getting the first derivative? Please show what you've tried.

5. Here's what I got for the 1st and 2nd derivative:

f'(x) = [be^(-1/2 b^2 (a-x)^2)] [a-x] / sqrt[2*pi]

f"(x) = [be^(-1/2 b^2 (a-x)^2)] [a^2 b^2 + x^2 b^2 -2axb^2 - 1] / sqrt [2 *pi]

Too many variables. I don't know if derivatives are correct or how to solve for 0.

6. Originally Posted by mat220-09
Here's what I got for the 1st and 2nd derivative:

f'(x) = [be^(-1/2 b^2 (a-x)^2)] [a-x] / sqrt[2*pi]

f"(x) = [be^(-1/2 b^2 (a-x)^2)] [a^2 b^2 + x^2 b^2 -2axb^2 - 1] / sqrt [2 *pi]

Too many variables. I don't know if derivatives are correct or how to solve for 0.
I would introduce new symbols: $f(x) = A e^{-k(x - a)^2}$

The differentiations are less cumbersome and it's easier to see how to factorise and solve the equation $f''(x) = 0$. Substitute as required the old symbols once you've solved for x.

Note that learning latex will make your posts easier to read. This will probably increase the pool of people willing to reply, which will lead to getting faster replies.